Question

When a potential difference V is applied across a wire of resistance R, it dissipates energy at a rate W. If the wire is cut into two halves and these halves are connected mutually parallel across the same supply, the same supply, the energy dissipation rate will become:

• A. $\frac{1}{4}W$
• B. $\frac{1}{2}W$
• C. 4W
• D. 2W

Answer 1

The Correct Option is C

To solve this problem, we need to analyze the energy dissipation rate before and after the wire is cut.

Initial Energy Dissipation Rate: The power P dissipated by a resistor is given by:

\( P = \frac{V^2}{R} \),

where: - V is the voltage across the resistor, and R is the resistance.

Initially, with resistance R, the energy dissipation rate is:

\( W = P = \frac{V^2}{R} \).

After Cutting the Wire: When the wire is cut into two halves, each half has a resistance of:

\( R' = \frac{R}{2} \).

Connecting in Parallel: When these two halves are connected in parallel, the equivalent resistance R_eq is given by:

\( \frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'} = \frac{2}{R/2} = \frac{4}{R} \).

Thus, the equivalent resistance is:

\( R_{eq} = \frac{R}{4} \).

New Energy Dissipation Rate: The new power P' dissipated in the circuit with the new resistance R_eq is:

\( P' = \frac{V^2}{R_{eq}} = \frac{V^2}{R/4} = \frac{4V^2}{R} \).

Comparing Power Dissipation Rates: Since \( W = \frac{V^2}{R} \) from the original circuit, we can relate the new power:

\( P' = 4W \).